Quick Check

From the signatures, we can tell that they are different:

pow(x, y[, z])

math.pow(x, y)

Also, trying it in the shell will give you a quick idea:

>>> pow is math.pow
False

Testing the differences

Another way to understand the differences in behaviour between the two functions is to test for them:

import math
import traceback
import sys

inf = float("inf")
NaN = float("nan")

vals = [inf, NaN, 0.0, 1.0, 2.2, -1.0, -0.0, -2.2, -inf, 1, 0, 2]

tests = set([])

for vala in vals:
  for valb in vals:
    tests.add( (vala, valb) )
    tests.add( (valb, vala) )


for a,b in tests:
  print("math.pow(%f,%f)"%(a,b) )
  try:
    print("    %f "%math.pow(a,b))
  except:
    traceback.print_exc()
  
  print("__builtins__.pow(%f,%f)"%(a,b) )
  try:
    print("    %f "%__builtins__.pow(a,b))
  except:
    traceback.print_exc()

We can then notice some subtle differences. For example:

math.pow(0.000000,-2.200000)
    ValueError: math domain error

__builtins__.pow(0.000000,-2.200000)
    ZeroDivisionError: 0.0 cannot be raised to a negative power

There are other differences, and the test list above is not complete (no long numbers, no complex, etc...), but this will give us a pragmatic list of how the two functions behave differently. I would also recommend extending the above test to check for the type that each function returns. You could probably write something similar that creates a report of the differences between the two functions.

math.pow()

math.pow() handles its arguments very differently from the builtin ** or pow(). This comes at the cost of flexibility. Having a look at the source, we can see that the arguments to math.pow() are cast directly to doubles:

static PyObject *
math_pow(PyObject *self, PyObject *args)
{
    PyObject *ox, *oy;
    double r, x, y;
    int odd_y;

    if (! PyArg_UnpackTuple(args, "pow", 2, 2, &ox, &oy))
        return NULL;
    x = PyFloat_AsDouble(ox);
    y = PyFloat_AsDouble(oy);
/*...*/

The checks are then carried out against the doubles for validity, and then the result is passed to the underlying C math library.

builtin pow()

The built-in pow() (same as the ** operator) on the other hand behaves very differently, it actually uses the Objects's own implementation of the ** operator, which can be overridden by the end user if need be by replacing a number's __pow__(), __rpow__() or __ipow__(), method.

For built-in types, it is instructive to study the difference between the power function implemented for two numeric types, for example, floats, long and complex.

Overriding the default behaviour

Emulating numeric types is described here. essentially, if you are creating a new type for numbers with uncertainty, what you will have to do is provide the __pow__(), __rpow__() and possibly __ipow__() methods for your type. This will allow your numbers to be used with the operator:

class Uncertain:
  def __init__(self, x, delta=0):
    self.delta = delta
    self.x = x
  def __pow__(self, other):
    return Uncertain(
      self.x**other.x, 
      Uncertain._propagate_power(self, other)
    )
  @staticmethod
  def _propagate_power(A, B):
    return math.sqrt(
      ((B.x*(A.x**(B.x-1)))**2)*A.delta*A.delta +
      (((A.x**B.x)*math.log(B.x))**2)*B.delta*B.delta
    )

In order to override math.pow() you will have to monkey patch it to support your new type:

def new_pow(a,b):
    _a = Uncertain(a)
    _b = Uncertain(b)
    return _a ** _b

math.pow = new_pow

Note that for this to work you'll have to wrangle the Uncertain class to cope with an Uncertain instance as an input to __init__()

The easy way is replace! One simple example:

value=str('6,0865000000e-01')
value2=value.replace(',', '.')
float(value2)
0.60865000000000002

One way to avoid string conversions is to implement the methods using Decimals:

from decimal import Decimal

def fexp(number):
    (sign, digits, exponent) = Decimal(number).as_tuple()
    return len(digits) + exponent - 1

def fman(number):
    return Decimal(number).scaleb(-fexp(number)).normalize()

Note that using floating point numbers, its not possible to calculate mantissa and exponent without rounding. The reason is that floating point numbers are stored as base 2 fractions. For example stored float value for 154.3 is 154.30000000000001136868377216160297393798828125. Floats are displayed in console as accurate numbers, because (in CPython) they are always rounded when serialized using a hard-coded precision of 17.

There is a no way to control that, best way is to write a function for this e.g.

def eformat(f, prec, exp_digits):
    s = "%.*e"%(prec, f)
    mantissa, exp = s.split('e')
    # add 1 to digits as 1 is taken by sign +/-
    return "%se%+0*d"%(mantissa, exp_digits+1, int(exp))

print(eformat(0.0000870927939438012, 14, 3))
print(eformat(1.0000870927939438012e5, 14, 3))
print(eformat(1.1e123, 4, 4))
print(eformat(1.1e-123, 4, 4))

Output:

8.70927939438012e-005
1.00008709279394e+005
1.1000e+0123
1.1000e-0123

Nothing to do with exponent. Problem is comma instead of decimal point.

>>> float("6,52353753563E-7")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for float(): 6,52353753563E-7
>>> float("6.52353753563E-7")
6.5235375356299998e-07

For a general approach, see locale.atof()